The function,Originally Posted by c_323_h
is countinous on
By IVT,
For any there exists an such as,
Let and the proof is finished.
Use the Intermediate Value Theorem to show that has a solution.
Given:
and
I understand what the IVT is and how to find a approximate solution using trial and error, but what exactly does it mean by "show?" Are they asking for a proof? I understand the concept of IVT but I can't translate my potential way of solving this into math symbols and such
Thanks
This is what I tried. Let . Then, by the IVT, there exists a number such thatOriginally Posted by ThePerfectHacker
. Since the function is continuous over the interval there exists a number such that and
I don't know if that proves anything though. I really don't have a direction when I try to prove things. I just copied the definition of IVT out of the book.
Let me try this again.
If is a real function countinous on closed interval , and let be any number between and then there must exist an on such as, .
Root location theorem: If is countinous on and (in other words opposite signs) then there is a solution to the equation on
Proof: By the condition we have that and have opposite signs. Thus, zero is a number between them. Thus, by the intermediate value theorem, for some on
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Returning to your problem.
1)Given on
2)The function is countinous cuz all polynomials are.
3)And, and
4)And,
5)Condition for Root location theorem are satisfied.
6)Thus there exists a solution to .